Question

(7/12)power a+3/2 ÷(7/12)power-7a+4 = (7/12)power3a÷(7/12)power4a+6​

Answer 1

$\frac{(\frac{7}{12})^{a + \frac{3}{2} } }{(\frac{7}{12})^{-7a + 4} } = \frac{(\frac{7}{12})^{3a } }{(\frac{7}{12})^{4a + 6} }$

as the bases(7/12) are same we can simply cancel the bases and take only the powers.

$\frac{a + \frac{3}{2} }{-7a + 4} = \frac{3a }{4a + 6}$

$(a + \frac{3}{2} ) ( 4a + 6) = (3a) (-7a +6)$

$(\frac{2a + 3}{2}) ( 4a + 6) = (3a) (-7a +6)$

$(2a +3) (2a + 3) = (3a) (-7a +6)$

$(2a +3)^{2} = (3a) (-7a +6)$

$4a^{2} + 9 + 12a = -21a + 12a$

$4a^{2} + 9 + 21^{2} = 0$    { cancelling 12a on both sides}

$25a^{2} = -9$

$a^{2} = \frac{-9}{25}$

$a = \sqrt{\frac{9}{25} } or -\sqrt{(\frac{9}{25}) }$

$a =\frac{3}{5} or- \frac{3}{5}$

HOPE THIS ANSWER HELPS YOU.

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